Unit-speed reparametrization on $a(t)=(\cosh (t),\sinh (t),t)$

Question

Well, at first we can get an arc length function of $a(t)$,

say $s(t)=\sqrt2 \sinh t$

Then $\frac{ds}{dt}=\sqrt2\cosh t$

Then we want to get $t=t(s)$ whose derivatives $\frac{dt}{ds}$ at $s=s(t)$ is the reciprocal of $\frac{ds}{dt}$ at $t=t(s).$

So, $\frac{dt}{ds}$ = $\frac{1}{\sqrt2 \cosh(s)}$

Then by integral, we can get $t(s)$, but it is wrong.

I think $\frac{dt}{ds}$ was wrong, but I cannot understand what is wrong.

Answer 1

The parameter $t = t(s)$ is the inverse of the arc length function, so you need to calculate $s^{-1}(t)$. Then the reparametrization is $\tilde{\gamma}(s) = (\gamma \circ t)(s)$. This is unit speed since by the chain rule,

$$\frac{d\tilde{\gamma}}{ds} = \frac{d\gamma}{dt} \cdot \frac{dt}{ds} = \frac{d\gamma}{dt} \cdot \frac{1}{ds/dt} = \frac{\gamma'}{\|\gamma'\|} $$